Question

The solution of the differential equation $\frac{dy}{dx}+\frac{1+x^2}{x}=0$ is

Answer 1

$\frac{dy}{dx}=\frac{-1-x^2}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{x}-\frac{x^2}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{-1}{x}-x$

$\Rightarrow dy=\frac{-dx}{x}-xdx$

$\Rightarrow \int dy=\int \frac{-dx}{x}-\int xdx$

$\Rightarrow \frac{y^2}{2}=-\ln x-\frac{x^2}{2}+c_1$

$\Rightarrow \frac{y^2}{2}+\frac{x^2}{2}=-\ln x+c_1$

$\Rightarrow x^2+y^2=-2\ln x+2c_1$

$\therefore x^2+y^2=-2\ln x+c$ where $c=2c_1$